5
Corollary 2.6
Let
M
be a faithful multiplication
R
-module and
N
a submodule of
M
. Then
N
is co-essential in
M
if
and only if (
N
:
M
) is co-essential in
R
.
Proposition 2.7
Let
M
be a faithful multiplication
R
-module,
N
a submodule of
M
and
I
an ideal of
R
. Then the
following statements are satisfied.
(1) If
N
is co-essential in
M
, then
IN
is co-essential in
M
.
(2) If
I
is co-essential in
R
, then
IN
is co-essential in
M
.
Proof
(1) The proof is straightforward.
(2) Suppose that
I
is co-essential in
R
. Let
X
be a submodule of
M
such that
IN + X = M
.
Since
M
is a
multiplication module,
N
= (
N
:
M
)
M
and
X
= (
X
:
M
)
M
. Thus,
M
=
IN + X
= [
I
(
N
:
M
)
+
(
X
:
M
)]
M
[
I +
(
X
:
M
)]
M
M
,
so that [
I +
(
X
:
M
)]
M
=
M
. By Lemma 0.3 (2),
I +
(
X
:
M
) =
R
and hence (
X
:
M
) =
R
since
I
is co-essential in
R
.
Therefore,
X
=
M
and
IN
is co-essential in
M
.
□
3.
Uniform
and co-uniform submodules
In this section we proof some proposition concerned with relationships between uniform resp. co-uniform
ideals of a ring
R
and uniform resp. co-uniform modules submodules of a multiplication
R
-module
M
. Let us recall that
an
R
-module
M
is said to be
uniform module
if the intersection of any two non-zero submodules of
M
is non-zero, [10].
An
R
-module
M
is said to be
co-uniform
if for every proper submodule of
M
is co-essential
in
M
, [1].
Proposition 3.1
Let
M
be a faithful multiplication
R
-module. Then a submodule
N
of
M
is uniform if and only if there
exists uniform ideal
I
of
R
such that
N
=
IM
.
Proof
Suppose that
N
is a uniform submodule of
M
. Then there exists an ideal
I
of
R
such that
N
=
IM
. It is
sufficient to prove that
I
is uniform. For any non-zero ideal
A
and
B
of
I
, we have
AM
and
BM
are non-zero submodule
of
N
. Thus, (
A
B
)
M
=
AM
BM
0 by hypothesis. Hence,
A
B
0 and we have
I
is a uniform ideal of
R
.
Conversely, suppose that
N
=
IM
for some uniform ideal
I
of
R
. Let
V
and
W
be non-zero submodules of
N
. By
Lemma 0.1 (2), we can choose non-zero subideals
A
and
B
of
I
such that
V
=
AM
and
W
=
BM
. Thus,
V
W
=
AM
BM =
(
A
B
)
M
again by Lemma 0.1 (1). By hypothesis,
A
B
0 and hence
V
W
0 because
M
is a
faithful. This implies that
N
is a uniform submodule of
M
. The proof is complete.
□
Corollary 3.2
Let
M
be a faithful multiplication
R
-module and
N
a submodule of
M
. Then
N
is a uniform submodule
of
M
if and only if (
N
:
M
) is a uniform ideal of
R
.
Proposition 3.3
Let
M
be a faithful multiplication
R
-module. Then
M
is co-uniform if and only if every proper ideal
of
R
is co-essential.
Proof
Suppose that
M
is co-uniform. Let
I
be a proper ideal of
R
such that
I + J = R
for every ideal
J
of
R
. Thus,
IM + JM = M
. Since
I
is a proper ideal of
R
and by Lemma 0.3 (3),
IM
M
. By hypothesis,
JM = M
and hence
J = R
again by Lemma 0.3 (2). Therefore,
I
is co-essential in
R
. Conversely, suppose that every proper ideal of
R
is
co-essential. Let
N
be a proper submodule of
M
such that
N + X = M
for every submodule
X
of
M
. Then
N
=
IM
and
